# RD SHARMA Solutions for Class 10 Maths Chapter 1 - Real Numbers

## Chapter 1 - Real Numbers Exercise Ex. 1.1

**For any positive integer n, prove that n ^{3} - n divisible by 6.**

**Prove that the square of any positive integer is of the form 5q, 5q + 1, 5q + 4 for some integer q.**

**Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.**

**Let a be any odd positive integer we need to prove that a is of the form 6q+1 , or 6q+3 , or 6q+5 , where q is some integer.****Since a is an integer consider b = 6 another integer applying Euclid's division lemma we get ****a = 6q + r f or some integer q 0, and r = 0, 1, 2, 3, 4, 5 since ****0 r < 6.****Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5****However since a is odd so a cannot take the values 6q, 6q+2 and 6q+4****(since all these are divisible by 2) ****Also, 6q + 1 = 2 x 3q + 1 = 2k1 + 1, where k1 is a positive integer****6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer****6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer****Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.****Therefore, 6q + 1, 6q + 3, 6q + 5 are odd numbers. ****Therefore, any odd integer can be expressed is of the form ****6q + 1, or 6q + 3, or 6q + 5 where q is some integer****Concept Insight: In order to solve such problems Euclid's division lemma is applied to two integers a and b the integer b must be taken in accordance with what is to be proved, for example here the integer b was taken 6 because a must be of the form 6q + 1, 6q + 3, 6q + 5.****Basic definition of even (divisible by 2) and odd numbers (not divisible by 2) and the fact that addition and multiplication of integers is always an integer are applicable here.**

Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

[NCERT EXEMPLER]

Let x be any positive integer.

When we divide x by 6, the remainder is either 0 or 1 or 2 or 3 or 4 or 5.

So, x can be written as

x = 6a or x = 6a + 1 or x = 6a + 2 or x = 6a + 3 or x = 6a + 4 or x = 6a + 5.

Thus, we have the following cases:

__CASE I:__

When x = 6a,

x^{2} = 36a^{2} =
6(6a^{2}) = 6m, where m = 6a^{2}

__CASE II:__

When x = 6a + 1,

x^{2} = (6a + 1)^{2}
= 36a^{2} + 12a + 1 = 6(6a^{2} + 2a) + 1 = 6m + 1, where m =
6a^{2} + 2a

__CASE III:__

When x = 6a + 2,

x^{2} = (6a + 2)^{2}
= 36a^{2} + 24a + 4 = 6(6a^{2} + 4a) + 4 = 6m + 4, where m =
6a^{2} + 4a

__CASE IV:__

When x = 6a + 3,

x^{2} = (6a + 3)^{2}
= 36a^{2} + 36a + 9 = (36a^{2} + 36a + 6) + 3 = 6(6a^{2}
+ 6a + 1) + 3 = 6m + 3, where m = 6a^{2} + 6a + 1

__CASE V:__

When x = 6a + 4,

x^{2} = (6a + 4)^{2}
= 36a^{2} + 48a + 16 = (36a^{2} + 48a + 6) + 10 = 6(6a^{2}
+ 8a + 1) + 10 = 6m + 10, where m = 6a^{2} + 8a + 10

__CASE VI:__

When x = 6a + 5,

x^{2} = (6a + 5)^{2}
= 36a^{2} + 60a + 25 = (36a^{2} + 60a + 6) + 19 = 6(6a^{2}
+ 10a + 1) + 19 = 6m + 19, where m = 6a^{2} + 10a + 19

Here, x is of the form 6m or 6m + 1 or 6m + 3 or 6m + 4 or 6m + 10 or 6mn + 19.

So, it cannot be of the form 6m + 2 or 6m + 5.

Show that the cube of a positive integer is of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5.

Let x be any positive integer.

Then, it is of the form 6n, 6n + 1, 6n + 2, 6n + 3, 6n + 4 or 6n + 5.

So, we have the following cases:

__CASE I:__

When x = 6n,

x^{3} = (6n)^{3} =
216n^{3} = 6(36n^{3}) = 6q, where q = 36n^{3}

__CASE II:__

When x = 6n + 1,

x^{3} = (6n + 1)^{3}
= 216n^{3} + 108n^{2} + 18n + 1 = 6(36n^{3} + 18n^{2}
+ 3n) + 1 = 6q + 1, where q = 36n^{3} + 18n^{2} + 3n

__CASE III:__

When x = 6n + 2,

x^{3} = (6n + 2)^{3}
= 216n^{3} + 216n^{2} + 72n + 8 = 216n^{3} + 216n^{2}
+ 72n + 6 + 2 =6(36n^{3} + 36n^{2} + 12n + 1) + 2 = 6q + 2,
where q = 36n^{3} + 54n^{2} + 12n + 1

__CASE IV:__

When x = 6n + 3,

x^{3} = (6n + 3)^{3}
= 216n^{3} + 324n^{2} + 162n + 27 = 216n^{3} + 324n^{2}
+ 162n + 24 + 3 =6(36n^{3} + 54n^{2} + 27n + 4) + 3 = 6q + 3,
where q = 36n^{3} + 54n^{2} + 27n + 4

__CASE V:__

When x = 6n + 4,

x^{3} = (6n + 4)^{3}
= 216n^{3} + 432n^{2} + 288n^{2} + 64 = 216n^{3}
+ 432n^{2} + 288n + 60 + 4 =6(36n^{3} + 72n^{2} + 48n
+ 10) + 4 = 6q + 4, where q = 36n^{3} + 72n^{2} + 48n + 10

__CASE VI:__

When x = 6n + 5,

x^{3} = (6n + 5)^{3}
= 216n^{3} + 540n^{2} + 450n^{2} + 125 = 216n^{3}
+ 540n^{2} + 450n + 120 + 5 =6(36n^{3} + 90n^{2} + 75n
+ 20) + 5 = 6q + 5, where q = 36n^{3} + 72n^{2} + 48n + 10

Thus, the cube of any positive integer is of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5.

Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.

Given numbers are n, n + 4, n + 8, n + 12 and n + 16.

Let n = 5q + r, where 0 ≤ r < 5

n = 5q, 5q + 1, 5q + 2, 5q + 3 or 5q + 4 for any natural number q.

So, we have the following cases:

__CASE I:__

When n = 5q

**n = 5q is divisible by 5**

n + 4 = 5q + 4 is not divisible by 5

n + 8 = 5q + 8 = 5q + 5 + 3 = 5(q + 1) + 3 is not divisible by 5

n + 12 = 5q + 12 = 5q + 10 + 2 = 5(q + 2) + 2 is not divisible by 5

n + 16 = 5q + 16 = 5q + 15 + 1 = 5(q + 3) + 1 is not divisible by 5

__CASE II:__

When n = 5q + 1

n = 5q + 1 is not divisible by 5

**n + 4 = 5q + 1 + 4 = 5q + 5 =
5(q + 1) is divisible by 5**

n + 8 = 5q + 1 + 8 = 5q + 9 = 5q + 5 + 4 = 5(q + 1) + 4 is not divisible by 5

n + 12 = 5q + 1 + 12 = 5q + 13 = 5q + 10 + 3 = 5(q + 2) + 3 is not divisible by 5

n + 16 = 5q + 1 + 16 = 5q + 17 = 5q + 15 + 2 = 5(q + 3) + 2 is not divisible by 5

__CASE III:__

When n = 5q + 2

n = 5q + 2 is not divisible by 5

n + 4 = 5q + 2 + 4 = 5q + 6 = 5q + 5 + 1 = 5(q + 1) + 1 is not divisible by 5

**n + 8 = 5q + 2 + 8 = 5q + 10 =
5(q + 2) is divisible by 5**

n + 12 = 5q + 2 + 12 = 5q + 14 = 5q + 10 + 4 = 5(q + 2) + 4 is not divisible by 5

n + 16 = 5q + 2 + 16 = 5q + 18 = 5q + 15 + 3 = 5(q + 3) + 3 is not divisible by 5

__CASE IV:__

When n = 5q + 3

n = 5q + 3 is not divisible by 5

n + 4 = 5q + 3 + 4 = 5q + 7 = 5q + 5 + 2 = 5(q + 1) + 2 is not divisible by 5

n + 8 = 5q + 3 + 8 = 5q + 11 = 5(q + 2) + 1 is not divisible by 5

**n + 12 = 5q + 3 + 12 = 5q + 15
= 5(q + 3) is divisible by 5**

n + 16 = 5q + 3 + 16 = 5q + 19 = 5q + 15 + 4 = 5(q + 3) + 4 is not divisible by 5

__CASE V:__

When n = 5q + 4

n = 5q + 4 is not divisible by 5

n + 4 = 5q + 4 + 4 = 5q + 8 = 5q + 5 + 3 = 5(q + 1) + 3 is not divisible by 5

n + 8 = 5q + 4 + 8 = 5q + 12 = 5(q + 2) + 2 is not divisible by 5

n + 12 = 5q + 4 + 12 = 5q + 16 = 5q + 15 + 1 = 5(q + 3) + 1 is not divisible by 5

**n + 16 = 5q + 4 + 16 = 5q + 20
= 5(q + 4) is divisible by 5**

Hence, in each case, one and only one out of n, n + 4, , n + 8, n + 12 and n + 16 is divisible by 5.

Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.

We know that any positive integer can be of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5 for some integer m.

Thus, an odd positive integer x can be of the form 6m + 1, 6m + 3 or 6m + 5.

Thus, we have

__CASE I:__

x = 6m + 1

x^{2} = (6m + 1)^{2}
= 36m^{2} + 12m + 1 = 6(6m^{2} + 2m) + 1 = 6q + 1, where q =
6m^{2} + 2m

__CASE II:__

x = 6m + 3

x^{2} = (6m + 3)^{2}
= 36m^{2} + 36m + 9 = 36m^{2} + 36m + 6 + 3 = 6(6m^{2}
+ 6m + 1) + 3 = 6q + 3, where q = 6m^{2} + 6m + 1

__CASE III:__

x = 6m + 5

x^{2} = (6m + 5)^{2}
= 36m^{2} + 60m + 25 = 36m^{2} + 60m + 24 + 1 = 6(6m^{2}
+ 10m + 4) + 1 = 6q + 1, where q = 6m^{2} + 10m + 4

Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.

A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any form other than 3m + 1, 3m or 3m + 2 for some integer m? Justify your answer.

By Euclid's Lemma,

a = bq + r, 0 ≤ r < b

Here, a is any positive integer and b = 3,

a = 3q + r

So, this must be in the form 3q, 3q + 1 or 3q + 2.

Now,

(3q)^{2} = 9q^{2} =
3m

(3q + 1)^{2} = 9q^{2}
+ 6q + 1 = 3(3q^{2} + 2q) + 1 = 3m + 1

(3q + 2)^{2} = 9q^{2}
+ 12q + 4 = 9q^{2} + 12q + 3 + 1 = 3(3q^{2} + 4q + 1) + 1 =
3m + 1

Thus, square of a positive integer of the form 3q + 1 is always of the form 3m + 1 or 3m for some integer m.

Show that the square of any positive integer cannot be of the form 3m + 2, where m is a natural number.

Let x be any positive integer.

So, x can be written as

x = 3a or x = 3a + 1 or x = 3a + 2

Thus, we have the following cases:

__CASE I:__

When x = 3a,

x^{2} = 9a^{2} =
3(3a^{2}) = 3m, where m = 3a^{2}

__CASE II:__

When x = 3a + 1,

x^{2} = (3a + 1)^{2}
= 9a^{2} + 6a + 1 = 3(3a^{2} + 2a) + 1 = 3m + 1, where m = 3a^{2}
+ 2a

__CASE III:__

When x = 3a + 2,

x^{2} = (3a + 2)^{2}
= 9a^{2} + 12a + 4 = 9a^{2} + 12a + 3 + 1 = 3(3a^{2}
+ 4a + 1) + 1 = 3m + 1, where m = 3a^{2} + 4a + 1

Thus, the square of any positive integer cannot be of the form 3m + 2, where m is a natural number.

## Chapter 1 - Real Numbers Exercise Ex. 1.2

**Find H.C.F. of 32 and 54**

Use Euclid's division algorithm to find the HCF of:

135 and 225

135 and 225

Step 1: Since 225 > 135, apply Euclid's division lemma, to a =225 and b=135 to find q and r such that 225 = 135q+r, 0 r

On dividing 225 by 135 we get quotient as 1 and remainder as 90

i.e 225 = 135 x 1 + 90

Step 2: Remainder r which is 90 0, we apply Euclid's division lemma to a = 135 and b = 90 to find whole numbers q and r such that

135 = 90 x q + r 0 r<90

On dividing 135 by 90 we get quotient as 1 and remainder as 45

i.e 135 = 90 x 1 + 45

Step 3: Again remainder r = 45 0 so we apply Euclid's division lemma to a = 90 and b = 45 to find q and r such that

90 = 45 x q + r 0 r<45

On dividing 90 by 45 we get quotient as 2 and remainder as 0

i.e 90 = 2 x 45 + 0

Step 4: Since the remainder is zero, the divisor at this stage will be HCF of (135, 225).

Since the divisor at this stage is 45, therefore, the HCF of 135 and 225 is 45.

Use Euclid's division algorithm to find the HCF of

184, 230 and 276

Use Euclid's division algorithm to find the HCF

136, 170 and 255

**Find H.C.F. of 592 and 252 and express it as a linear combination of them.**

**Find the largest number which divides 615 and 963 leaving remainder 6 in each case.**

Using Euclid's division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively.

**105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went in each trip?**

Use Euclid's division algorithm to find the HCF of:

196 and 38220

196 and 38220

Step 1: Since 38220 > 196, apply Euclid's division lemma

to a =38220 and b=196 to find whole numbers q and r such that

38220 = 196 q + r, 0 r < 196

On dividing 38220 we get quotient as 195 and remainder r as 0

i.e 38220 = 196 x 195 + 0

Since the remainder is zero, divisor at this stage will be HCF

Since divisor at this stage is 196 , therefore, HCF of 196 and 38220 is 196.

NOTE: HCF( a,b) = a if a is a factor of b. Here, 196 is a factor of 38220 so HCF is 196.

Use Euclid's division algorithm to find the HCF of:

867 and 255

867 and 255

Step 1: Since 867 > 255, apply Euclid's division lemma, to a =867 and b=255 to find q and r such that 867 = 255q + r, 0 r<255

On dividing 867 by 255 we get quotient as 3 and remainder as 102

i.e 867 = 255 x 3 + 102

Step 2: Since remainder 102 0, we apply the division lemma to a=255 and b= 102 to find whole numbers q and r such that

255 = 102q + r where 0 r<102

On dividing 255 by 102 we get quotient as 2 and remainder as 51

i.e 255 = 102 x 2 + 51

Step 3: Again remainder 51 is non zero, so we apply the division lemma to a=102 and b= 51 to find whole numbers q and r such that

102 = 51 q + r where 0 r < 51

On dividing 102 by 51 quotient is 2 and remainder is 0

i.e 102 = 51 x 2 + 0

Since the remainder is zero, the divisor at this stage is the HCF

Since the divisor at this stage is 51,therefore, HCF of 867 and 255 is 51.

Concept Insight: To crack such problem remember to apply the Euclid's division Lemma which states that "Given positive integers a and b, there exists unique integers q and r satisfying a = bq + r, where 0 r < b" in the correct order.

Here, a > b.

Euclid's algorithm works since Dividing 'a' by 'b', replacing 'b' by 'r' and 'a' by 'b' and repeating the process of division till remainder 0 is reached, gives a number which divides a and b exactly.

i.e HCF(a,b) =HCF(b,r)

Note that do not find the HCF using prime factorisation in this question when the method is specified and do not skip steps.

lemma to a=135 and b=

## Chapter 1 - Real Numbers Exercise Ex. 1.3

**Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.**

**Numbers are of two types - prime and composite. Prime numbers has only two factors namely 1 and the number itself whereas composite numbers have factors other than 1 and itself.**

**It can be observed that**

**7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) = 13 x (77 + 1)**

**= 13 x 78**

**= 13 x 13 x 6**

**The given expression has 6 and 13 as its factors. Therefore, it is a composite number.**

**7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1)**

**= 5 x (1008 + 1)**

**= 5 x 1009**

**1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.**

**Concept Insight: Definition of prime numbers and composite numbers is used. Do not miss the reasoning.**

**Check whether 6**

^{n}can end with the digit 0 for any natural number n.**If any number ends with the digit 0, it should be divisible by 10 or in other words its prime factorisation must include primes 2 and 5 both**

**Prime factorisation of 6**

^{n}= (2 x 3)^{n}**By Fundamental Theorem of Arithmetic Prime factorisation of a number is unique. So 5 is not a prime factor of 6**

^{n}.**Hence, for any value of n, 6n will not be divisible by 5.**

**Therefore, 6**

^{n}cannot end with the digit 0 for any natural number n.**Concept Insight: In order solve such problems the concept used is if a number is to end with zero then it must be divisible by 10 and the prime factorisation of a number is unique.**

Explain why 3 × 5 × 7 + 7 is a composite number.

Numbers are of two types - prime and composite. Prime numbers has only two factors namely 1 and the number itself whereas composite numbers have factors other than 1 and itself. It can be observed that 3 × 5 × 7+ 7 = 7 × (3 × 5 + 1) = 7 × (15 + 1) = 7 × 16

The given expression has 7 and 16 as its factors. Therefore, it is a
composite number.** **

## Chapter 1 - Real Numbers Exercise Ex. 1.4

**Find the LCM and HCF of the following integers by applying the prime factorisation method:**

**(i) 12,15 and 21**

**(ii) 17, 23 and 29**

**(iii) 8, 9 and 25**

**(iv) 40, 36 and 126**

**(v) 84, 90 and 120**

**(vi) 24, 15 and 36.**

**Concept Insight:**HCF is the product of common prime factors of all three numbers raised to least power, while LCM is product of prime factors of all here raised to highest power. Use the fact that HCF is always a factor of the LCM to verify the answer. Note HCF of (a,b,c) can also be calculated by taking two numbers at a time i.e HCF (a,b) and then HCF (b,c) .

**Given that HCF (306, 657) = 9, find LCM (306, 657).**

**Concept Insight: **This problem must be solved using product of two numbers = HCF x LCM rather than prime factorisation

**Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive).**

## Chapter 1 - Real Numbers Exercise Ex. 1.5

Prove that is an irrational number.

## Chapter 1 - Real Numbers Exercise Ex. 1.6

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Write the denominator of the rational number in the form 2^{m}
× 5^{n}, where m, n are non-negative
integers. Hence, write the decimal expansion, without actual division.

A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of q, when this number is expressed in the form? Give reasons.

## Chapter 1 - Real Numbers Exercise 1.59

The exponent of 2 in the prime factorisation of 144, is

(a) 4

(b) 5

(c) 6

(d) 3

Factorisation of 144 can be done as follows

so 144 = 2 × 2 × 2 × 2 × 3 × 3

= 2^{4} × 3^{2}

Exponent of 2 in the prime factorisation of 144 is 4.

So, the correct option is (a).

The LCM of two numbers is 1200. Which of the following cannot be their HCF ?

(a) 600

(b) 500

(c) 400

(d) 200

We know that LCM of two numbers is divisible of HCF of these two numbers.

Hence

(a) 1200 is divisible by 600. So 600 can be the HCF.

(b) 500 cannot be the HCF because 1200 is not divisible by 500.

(c) 400 can be the HCF because 1200 is divisible by 400.

(d) 200 can be the HCF because 1200 is divisible by 200.

So, the correct option is (b).

If n = 2^{3 }× 3^{4} × 5^{4} × 7, then the number of consecutive zeros in n, where n is a natural number, is

(a) 2

(b) 3

(c) 4

(d) 7

n can also be written as

3^{4} × 2^{3} × 5^{3} × 5 × 7

3^{4} × (2 × 5)^{3} × 5 × 7

3^{4} × 5 × 7 × 10^{3}

exponent of 10 in n is 3.

Hence number of consecutive zeros in n is 3.

So, the correct option is (b).

The sum of the exponents of the prime factors in the prime factorisation of 196, is

(a) 1

(b) 2

(c) 4

(c) 6

Factorisation of 196 is

so 196 = 2 × 2 × 7 × 7

= 2^{2} × 7^{2}

exponent of 2 is 2

exponent of 7 is 2

Hence sum of exponents is 4.

So, the correct option is (c).

(a) 1

(b) 2

(c) 3

(d) 4

So, the correct option is (b).

(a) an even number

(b) an odd number

(c) an odd prime number

(d) a prime number

So, the correct option is (a).

If two positive integers a and b are expressible in the form a = pq^{2 }and b = p^{3}q ; p, q being prime numbers,

then LCM (a, b) is

(a) pq

(b)

(c)

(d)

LCM (a, b) is

LCM (a, b) = p × q × q × p^{2}

= p^{3}q^{2}

So, the correct option is (c).

In Q. no. 7, HCF (a, b) is

(a) pq

(b)

(c)

(d)

HCF (a, b) is

No further common division is possible

Hence HCF (a, b) = p × q

= pq

So, the correct option is (a).

## Chapter 1 - Real Numbers Exercise 1.60

If two positive numbers m and n are expressible in the form m = pq^{3} and n = p^{3}q^{2}, where p, q are prime numbers,

then HCF (m, n) =

(a) pq

(b) pq^{2}

(c) p^{3}q^{3}

(d) p^{2}q^{3}

HCF of m, n is

No further division is possible

Hence HCF is p × q^{2 }= pq^{2}

So, the correct option is (b).

If the LCM of a and 18 is 36 and the HCF of a and 18 is 2, then a =

(a) 2

(b) 3

(c) 4

(d) 1

We know,

LCM (a, b) × HCF (a, b) = a × b

So LCM (a, 18) × HCF (a, 18) = a × 18

36 × 2 = a × 18

a = 4

So, the correct option is (c).

The HCF of 95 and 152, is

(a) 57

(b) 1

(c) 19

(d) 38

HCF (95, 152)

No further common division is possible.

Hence HCF (95, 152) is 19.

So, the correct option is (c).

If HCF (26, 169) = 13, then LCM (26, 169) =

(a) 26

(b) 52

(c) 338

(d) 13

We know

LCM (a, b) × HCF (a, b) = a × b

so LCM (26, 169) × HCF (26, 169) = 26 × 169

So, the correct option is (c).

If a = 2^{3} × 3, b = 2 × 3 × 5, c = 3^{n} × 5 and LCM (a, b, c) = 2^{3} × 3^{2} × 5 then n =

(a) 1

(b) 2

(c) 3

(d) 4

LCM (a, b, c) is

LCM (a, b, c) = 3 × 2 × 5 × 2^{2} × 3^{n - 1}

= 2^{3} × 3^{n} × 5 ........(1)

given that

LCM (a, b, c) = 2^{3} × 3^{2} × 5 ........(2)

from (1) & (2)

n = 2

So, the correct option is (b).

(a) one decimal place

(b) two decimal places

(c) three decimal places

(d) four decimal places

So, the correct option is (d).

If p and q are co - prime numbers, then p^{2} and q^{2} are

(a) coprime

(b) not coprime

(c) even

(d) odd

If p and q are co-prime numbers then

HCF (p, q) = 1

After squaring the numbers, we get p^{2 }and q^{2 }

If two numbers have HCF = 1 then after squaring the numbers their HCF remains equal to 1.

Hence HCF (p^{2} , q^{2}) = 1

so p^{2} and q^{2} are co - prime numbers.

Ex : 2 and 3 are co - prime numbers.

HCF (2, 3) = 1

after squaring

HCF (4, 9) = 1

Hence, 4, 9 are also co - prime.

So, squares of two co - prime numbers are also co - prime.

So, the correct option is (a).

(a) (i) and (ii)

(b) (ii) and (iii)

(c) (i) and (iii)

(d) (i) and (iv)

So, the correct option is (d).

*Note: Since the book has a typo error, the question has been modified.

If 3 is the least prime factor of number a and 7 is the least prime factor of number b,

then least prime factor of a + b is

(a) 2

(b) 3

(c) 5

(d) 10

It is given that 3 is the least prime factor of number a so a can be 3 (least possible value)

It is given that 7 is the least prime factor of number b so least possible value of b is 7.

Hence a + b = 10 (least possible value)

prime factors of 10 are 2 and 5

Hence the least prime factor of a + b is 2.

So, the correct option is (a).

(a) an integer

(b) a rational number

(c) a natural number

(d) an irrational number

We know that decimal expansion of a rational number is either terminating or non-terminating and recurring.

So the correct option is (b).

(a)

(b)

(c)

(d) 3

If n is a natural number, then 9^{2n} - 4^{2n }is always divisible by

(a) 5

(b) 13

(c) both 5 and 13

(d) None of these

We know a^{2n } - b^{2n }is always divisible by a - b and a + b

On comparing with 9^{2n} - 4^{2n}, we get a = 9 & b = 4

Hence 9^{2n} - 4^{2n } is divisible by 9 - 4 & 9 + 4

= 5 & 13

So, the correct option is (c).

## Chapter 1 - Real Numbers Exercise 1.61

If n is any natural number, then 6^{n } - 5^{n } always ends with

(a) 1

(b) 3

(c) 5

(d) 7

6^{n }always ends with 6

5^{n} always ends with 5

Hence 6^{n }- 5^{n} always with 6 - 5 = 1

So, the correct option is (a).

The LCM and HCF of two rational numbers are equal, then the numbers must be

(a) prime

(b) co-prime

(c) composite

(d) equal

(a) If two numbers are prime then their HCF must be 1 but LCM can't be 1

Example: 2, 3

LCM (2, 3) = 6

HCF (2, 3) = 1

(b) If two numbers are co - prime then their HCF must be 1 but LCM can't be 1.

(c) If two numbers are composite then their LCM and HCF can only be equal if the two numbers are same.

(d) If the numbers are equal.

Example: 6, 6

LCM (6, 6) = 6

HCF (6, 6) = 6

LCM = HCF

So, the correct option is (d).

If sum of LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF, then the product of two numbers is

(a) 203400

(b) 194400

(c) 198400

(d) 205400

Let numbers be a, b

It is given that LCM (a, b) + HCF (a, b) = 1260 ..........(1)

LCM (a, b) - HCF (a, b) = 900 ..........(2)

Adding equations (1) and (2), we get 2LCM (a, b) = 2160

Subtracting equations (1) and (2), we get 2HCF (a, b) = 360

So, LCM (a, b) = 1080 and

HCF (a, b) = 180

We know LCM (a, b) × HCF (a, b) = ab

ab = 1080 × 180

= 194400

So, the correct option is (b).

The remainder when the square of any prime number greater than 3 is divided by 6, is

(a) 1

(b) 3

(c) 2

(d) 4

For some integer m, every even integer is of the form

- m
- m + 1
- 2m
- 2m + 1

m is an integer.

⇒ m = ….., -2, -1, 0, 1, 2, …..

⇒ 2m = ……., -4, -2, 0, 2, 4, ……

Hence, correct option is (c).

For some integer q, every odd integer is of the form

- q
- q + 1
- 2q
- 2q + 1

q is an integer.

⇒ q = ….., -2, -1, 0, 1, 2, …..

⇒ 2q + 1 = ……., -3, -1, 0, 3, 5, ……

Hence, correct option is (d).

n^{2} - 1 is divisible by 8, if n is

- an integer
- a natural number
- an odd integer
- an even integer

Let a = n^{2} - 1

Now, when n is odd, i.e. n = 2k + 1, we have

a = (2k + 1)^{2} - 1 =4k^{2}
+ 4k + 1 - 1 = 4k(k + 1)

At k = -1, we get

a = 4(-1)(-1 + 1) = 0, which is divisible by 8.

At k = 0, we get

a = 4(0)(0 + 1) = 0, which is divisible by 8.

At k = 1, we get

A = 4(1)(1 + 1) = 4(2) = 8, which is divisible by 8.

Hence, correct option is (c).

The decimal expansion of the rational number will terminate after

- one decimal place
- two decimal places
- three decimal places
- more than 3 decimal places

If two positive integers a and b are written as a =
x^{3}y^{2} and b = xy^{3}, x, y are prime numbers,
then HCF (a, b) is

- xy
- xy
^{2} - x
^{3}y^{3} - x
^{2}y^{2}

The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

- 10
- 100
- 504
- 2520

The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is

- 13
- 65
- 875
- 1750

If the HCF of 65 and 117 is expressible in the form 65m - 117, then the value of m is

- 4
- 2
- 1
- 3

Euclid's division lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy

- 1 < r < b
- 0 < r ≤ b
- 0 ≤ r < b
- 0 < r < b

Euclid's division lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy 0 ≤ r < b.

Hence, correct option is (c).

## Chapter 1 - Real Numbers Exercise 1.62

The decimal expansion of the rational number will terminate after:

- One decimal place
- Two decimal places
- Three decimal places
- Four decimal places

### Other Chapters for CBSE Class 10 Mathematics

Chapter 2- Polynomials Chapter 3- Pairs of Linear Equations in Two Variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- Trigonometric Identities Chapter 12- Heights and Distances Chapter 13- Areas Related to Circles Chapter 14- Surface Areas and Volumes Chapter 15- Statistics Chapter 16- Probability### RD SHARMA Solutions for CBSE Class 10 Subjects

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